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The difference of x^2 and x^3:Defining both x^2 and y^3 in terms of odd numbers that are shells of the power of 2 . . .
So when the difference of x^2 and y^3 is exactly "1" or when: The 1st through xth odd numbers equaling x^2, less some latter pth through qth odds equaling x^3, leaves "1", only the very first of the 1st through xth odds would qualify to be the remainder that is 1. Therefore, the x^3 value may ONLY begin from the second odd shell. So 2 (that is the 2nd shell) = ((y(y-1)/2) + 1) which means y = 2. If y = 2 then ((y(y+1))/2) = 3 which will be the top end of the range of odd shells that define this particular y^3. The 2nd odd plus the 3rd odd = 3 + 5 = 2^3 AND the sum of the 1st through 3rd shells of the power of 2 = 1 + 3 + 5 = 9 = 3^2. In the above way the Catalan conjecture is correct for squares and cubes . . . AND . . .
The Catalan conjecture is also true for n > 2:Eugene Charles Catalan's (1814-1894) conjecture that the only two powers whose difference would be 1 is: 3^2 - 2^3. In other words: For x^p - y^q = 1 x = 3, p = 2, y = 2 and q = 3 is the only answer.
For x^p and y^q (and z^n), if p and q and n are > 2, No x^p or y^q (or z^n) will be defined without including its very first shell of "1" so the difference of two terms at levels of power > 2 can never be 1. The logic is that the term that is the larger value will lose almost all of its value when the lesser value is subtracted, leaving 1--a 1st shell value. But, the 2nd through the zth of the larger value's shells which theoretically should have been the lesser term's total value, will not actually equal any viable definition of any z^n when n > 2. Definitions of z^n are determined by binomial theorem equations for shells that sum to z^n as well as by Euler/Pascal formulas for shell values. For all shells of powers, the first shell equals "1". For all powers, z^n = exclusively the 1st through the zth shells summed. See the Euler/Pascal cube of relations and Fermat's Last Theorem. If some theoretical y^q equals the 2nd to xth shells of the power of p, the "2nd to xth" shells is a contradiction, already impossible by definition if p > 2. (Notice that the one case of a solution of the formula occurs within the shells of the power of 2.) ###
Shell formulas will create x^n series by summing k = 1 through x: 1 = "Shell" formula S1 for x^1 2k – 1 = "Shell" formula S2 for x^2 3k^2 – 3k + 1 = "Shell" formula S3 for x^3 4k^3 – 6k^2 + 4k – 1 = "Shell" formula S4 for x^4 5k^4 – 10k^3 + 10k^2 – 5k + 1 = "Shell" formula S5 for x^5 6k^5 – 15k^4 + 20k^3 – 15k^2 + 6k – 1 = "Shell" formula S6 for x^6 etc. (Any x^n less 1 will leave the 2nd through xth shells of the shell series.)
The initial values of shell series look like this:
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